(navigation image)
Home American Libraries | Canadian Libraries | Universal Library | Community Texts | Project Gutenberg | Biodiversity Heritage Library | Children's Library | Additional Collections
Search: Advanced Search
Anonymous User (login or join us)
Upload
See other formats

Full text of "History Of The Theory Of Numbers - I"

264                    HISTORY OF THE THEORY OF NUMBERS.             [CHAP, rx
H. D. Carmichael10 treated the problem to find m, given the prime p and s =Soi, in Legendre's formula; a given solution m2 leads to an infinitude of solutions map*, k arbitrary. Again, to find m such that pm-< is the highest power of p>2 dividing ml, we have m t = (m  s)/(p  1), and see that w has a limited number of values; there is always at least one solution m.
Carmichael11 used the notation H\y\ for the index of the highest power of the prime p dividing y, and evaluated
U=o
where a, c are relatively prime positive integers. Set c0=c and let v be the least integer such that ia+Cr-i is divisible by p, the quotient being cr. Let
Then /i=2(er+l), where iis the least subscript for which
r-l
c,(a+c,)(2a+c<) . ..( is not divisible by p.   It follows that
where E is the index of the highest power of p not exceeding n  1 . If n is a power of p, /*= (n-l)/(p-l). But if n = 5kpk+ . . . +5^+^, 5A^0, and at least one further 8 is not zero,
. . .        . p-l          p~l
In case the first x for which xa+c is divisible by p gives c as the quotient, all the cr are equal and hence all the tr; then
, - [n  l-i+P
L     p
The case a = c = l yields Legendre's2 result.   The case a = 2, c = l, gi
gves
E. Stridsberg12 wrote JYm for (1) and considered
where a is any integer not divisible by the positive integer m. Let p be a prime not dividing m. Write a,- for the residue of aj modulo m. He noted that, if pj=l (mod m),
10Bull. Amer. Math. Soc., 14, 1907-8, 74-77; Arncr. Math. Monthly, 15, 1908, 15-17.
/Wd.f 15, 1908-9, 217.
lxArkiv for Matematik, Astr., Fyaik, 6, 1911, No. 34.